A beam of white light enters a transparent material. Wavelengths ...

Question

A beam of white light enters a transparent material. Wavelengths for which the index of refraction is n are refracted at angle θ₂. Wavelengths for which the index of refraction is n + δn, where δn << n, are refracted at angle θ₂ + δθ.b. A beam of white light is incident on a piece of glass at 30°. Deep violet light is refracted 0.28° more than deep red light. The index of refraction for deep red light is known to be 1.552. What is the index of refraction for deep violet light?

Accepted Answer

Hello, fellow physicists today, we're gonna solve the following practice problem together. So first off, let's read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. When a beam of light light passes through a transparent medium, it disperses into a spectrum of colors due to the varying refractive indices of different wavelengths consider a beam of light that is incident on a microscope slide with an angle of 20 degrees. It is observed that the wavelength lambda are nanometers having a refractive index. Nr 1.576 is refracted at an angle theta bar. The wavelength lambda B 470 nanometers having a refractive index NB is refracted at an angle theta R minus 0.32 degrees. Calculate NB hint when the refractive index varies, varies by partial N partial N is much smaller than N due to the wavelength variation. The angle of refraction varies by partial data. OK. So we're given some multiple choice answers. Let's read them off to see what our final answer might be. A is 1.540 B is 1.595 C is 1. and D is 1.725. OK. So first off, let us recall that the refractive index N varies as a function of the wavelength. Therefore, the angle of refraction varies with N to begin. Let us start by recalling and using the equation for Snell's law which states that N one multiplied by sine beta one is equal to N multiplied or mo and multiplied by sine beta two. OK. So not in this case, that N one is the refractive index of error. And that is 1.0 as its numerical value. So we can rewrite Snell's law as 1.0 multiplied by sine theta one. So 1.0 multiplied by sine theta one is equal to N multiplied by sine theta two. So we can rearrange Snell's law now to solve for theta two or I should say sign of theta two. And let's call that equation one. So sine of theta two is equal to sine theta one divided by N. So note that theta two represents the angle of refraction in the medium and that theta one represents the angle of incident, which is independent of N. Now we must take the partial derivative with respect to N. So partial divided by partial N of sine of theta two is equal to partial divided by partial N um sine theta one divided by N. So when we perform the partial derivative. We should get cosine theta to partial beta divided by partial N which is equal to negative sine beta one divided by N two, which is equal to negative N multiplied by sine theta two divided by N two. So equation two states that partial theta is equal to negative partial N divided by N multiplied by tangent. Theta two note that we're told that the wavelength lambda B is 470 nanometers has a refractive index of NB and is refracted at an angle theta R minus 0. degrees. Thus, we can write that partial theta is equal to negative 0.32 degrees. So now we need to use equation one to calculate the value for theta R. So theta R is equal to inverse sign of sine 20 degrees divided by 1. which equals 12. degrees. So tangent beta R equals 0.222 three. Now we need to rearrange equation two to calculate the partial end value. Let's do that. So partial N equals negative N divided by tangent of theta R multiplied by the partial of theta. So now we can plug in our known variables. So partial N is equal to negative 1.576 divided by 0. multiplied by negative 0. degrees multiplied by I multiplied by radiance divided by 180 degrees. So partial N is equal to 0. five. And that's when you plug all that into a calculator, that is the number you should see. So note that because 650 nanometers is longer than the 470 nanometer wavelength. So NR is less than NB. So we can write that the partial of N is equal to NB minus NR is equal to 0.039595. So when we rearrange to solve for NB, so NB is equal to NR plus 0. is equal to. And we know that Nr is 1. plus 0.039595. So that means when we add that into a calculator, NB should equal 1.612. And that is our final answer and this is the refractive index for NB. So let's look at our multiple choice answers to see what the correct answer should be. And it looks like the letter C is the correct answer one. So C is 1.612. Thank you so much for watching. Hopefully that helped and I can't wait to see you in the next video. Bye.

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